6 Answers
6

The easiest way (I know) to see that there are no nonconstant holomorphic maps from a complete elliptic curve $E$ to the stack $M_g$ is to observe that such a map $f$ would lift to a holomorphic map of the universal covers $\tilde{f}: {\mathbb C} \to T_g$, where $T_g$ is the Teichmuller space. The latter is a bounded domain in ${\mathbb C}^{3g-3}$, so Liouville's theorem implies that $f$ is constant.

Edit: Using Kodaira's construction of complete curves in moduli spaces (via ramified coverings of products of curves) one can construct maps from elliptic curves $E$ to the coarse moduli space (of large genus) which are generically 1-1, i.e. 1-1 away from a finite subset of $E$. With more work one can probably get injective maps as well but I do not see sufficient motivation for this.

If $C$ is a smooth curve of genus $g$ and $f:C\to M_g$ is a non-constant morphism to the stack, then the total space of the induced family of curves $S\to C$ is a smooth surface and the underlying oriented 4-manifold has non-trivial signature: it is given by a multiple of the pullback of the first Chern class of the Hodge bundle on $M_g$, which is an ample class. On the other hand, if a 4-manifold is a Riemann surface bundle over a torus or sphere, then its signature must be zero. Thus, there are no non-constant maps of an elliptic curve to the stack $M_g$. The question of what the smallest genus curve which maps non-trivially to $M_g$ is addressed in a paper I wrote with Ron Donagi around 10 years ago: http://arxiv.org/pdf/math.AG/0105203.pdf

I would guess that there does exist a non-constant map of an elliptic curve $C$ to the coarse space of $M_g$, but the map would only come from a family of curves after passing to a finite (ramified) cover of $C$. I think some of the examples constructed in the above paper are in fact of this form. I can think more about it if knowing the answer for the coarse space is important to you (but really, why would you be interested in the coarse space? :-) ).

Edit: I am assuming that the base field is $\mathbb{C}$ in the above answer. I don't know the answer for other fields.

Jim I love this: "why would you be interested in the coarse space?"!
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roy smithJul 26 '12 at 2:03

1

It was just pointed out to me by a friend that if you are studying the MMP program for $M_g$ (which is certainly an interesting endeavour), then you are more interested in the geometry of the coarse space. So maybe I need to walk back my flippant comment!
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Jim BryanJul 26 '12 at 3:33

1

Jim, if I understood their talks correctly, your old result was just used by Donagi and Witten to show that the moduli space of super-Riemann surfaces does not split -- meaning superstring perturbation theory itself needs to be "revisited."
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Eric ZaslowJul 26 '12 at 5:42

Does that mean that the moduli stack of super-Riemann surfaces might be split? If so, why do Donagi-Witten care whether the moduli space is split? (Sorry for getting off-topic.)
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Arend BayerJul 26 '12 at 12:42

@Jim: sometimes the coarse space is all you have. E.g., both the stack and the coarse space of $g$-dimensional ppav's have toroidal compactifications, but, for $g>1$, only the coarse space has a Satake compactification. It is this compactification that Faltings uses to prove the Mordell conjecture because it is here that there exists an ample line bundle with a good height function. More broadly, Mumford asks [GIT, ch. 5, l. 1] "What are moduli?", then gives no definitive answer, despite having already written the first paper to take stacks seriously as geometric objects.
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inkspotAug 2 '12 at 11:19

Over $\mathbb C$ this is a consequence of the Torelli theorem: A map from a rational or elliptic curve to $M_g$ is the same as a smooth family over that curve. Then considering the period map gives a map from the same curve to the parameter space of Hodge structures. However, that is hyperbolic, so any holomorphic map from $\mathbb C$ is trivial. Therefore the Hodge structures of the fibers are the same, but then Torelli says that then the curves are also the same, so the map to moduli is trivial. (In fact this proof shows that even a rational curve minus $2$ points cannot map there either).

Actually, a lot more is true:
The same statement holds if we replace $M_g$ with $M_h$, the moduli stack of canonically polarized smooth projective varieties with Hilbert polynomial $h$. (For $\deg h=1$ you get back the corresponding $M_g$). Torelli is no longer true, but the desired statement is: There are no non-trivial maps from an elliptic curve or a rational curve minus $2$ points. This is proved in Algebraic hyperbolicity of fine moduli spaces J. Algebraic Geom. 9 (2000), no. 1, 165–174.

If one wants to generalize further, that is, to include the case of rational curves minus two points, one possibility is

Viehweg's conjecture (roughly stated)
Any quasi-projective variety that admits a generically finite morphism to $M_h$ is of log general type.

Here log general type means that if $X$ is the variety in question and $\overline X$ is a projective variety such that $X\subseteq \overline X$ is an open subset and $D=\overline X\setminus X$ is a divisor, then $\omega_{\overline X}(D)$ is big (=has maximal Kodaira dimension).

If you have not seen this before, check that curves of log general type are:
Any open subset of a curve of genus at least $2$, any proper open subset of an elliptic curve, and any open subset of a rational curve missing at least $3$ points. In other words, the only non-log general type curves are the (projective) elliptic curves and rational curves missing at most $2$ points. In other words, Viehweg's conjecture for $M_g$ is just the first statement above.

As far as the coarse space goes, it is probably more of a curiosity, but it still seems interesting. Oort proved that the coarse space of $M_g$ actually contains rational curves. Perhaps his proof can be adapted to prove the same for an elliptic curve. (The main idea is to construct a family over a curve with a given map to $\mathbb P^1$ such that fibers over the points mapping to the same point on $\mathbb P^1$ are isomorphic, so the moduli map factors through the map to $\mathbb P^1$).

And then there are many results concerning the question of what kind of complete subvarieties might $M_g$ have. Or what about complete subvarieties through a general point? There are various results in this direction, but I am already digressing....

Remark all of the above holds over an algebraically closed field of characteristic $0$. In characteristic $p$ all kinds of weird things happen, so I would expect that probably most of this fails.

All these deductions on non-existence of an elliptic curve as in the question from the fact that $M_g$ is hyperbolic in various senses were very interesting. Here is a preposterous variant. If such an elliptic curve exists, then passing to a finitely generated field and specializing any transcendental (if necessary) we can assume that the elliptic curve and the embedding in $M_g$ are defined over a number field. We extend the number field so that the elliptic curve has infinitely many rational points over it. Now, it is easy to see that all these rational points (viewed as points in $M_g$) give rise to curves of genus $g$ defined over this number field and all having good reduction outside of a fixed finite set of places of the number field. This contradicts Shafarevich's conjecture (proved by Faltings).

Perhaps this can be made somewhat less preposterous using a function-field analogue of Shafarevich's conjecture? Often such analogues were proved somewhat earlier and more easily.
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Noam D. ElkiesJul 31 '12 at 14:41

But there are complete curves in $M_g$ (of genus bigger than one), so there are families of curves over function fields with good reduction everywhere, so the correct statement is bound to be subtle. These families are not deformable (Arakelov's theorem, which is the function field analogue of Shafarevich). Maybe using the fact that an elliptic curve has an infinite automorphism group gives a contradiction, but I am not sure.
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Felipe VolochJul 31 '12 at 15:36

3

Actually, what one may use to apply the function field version of Shafarevich is the fact that elliptic curves admit endomorphisms of degree larger than $1$. I included the details in an answer below. (I ran out of the space allowed for comments).
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Sándor KovácsAug 1 '12 at 0:00

Here is how the function field version of Shafarevich's conjecture (=Arakelov-Parshin Theorem) implies that there are no elliptic curves or (at most) twice punctured rational curves in $M_g$:
(See Noam's comment to Felipe's answer)

Suppose there exists a smooth non-isotrivial family $f:X\to C$ of curves of genus $g$ for some fixed $g>1$ parametrized by a curve $C$. Call such a family admissible, let $m\in\mathbb N$ fixed and consider the set of numbers
$$
D_m=\left\{
\deg (f_*\omega_{X/C}^m) \mid f \text{ is an admissible family }
\right\}
$$

By Shafarevich's conjecture (=Arakelov-Parshin Theorem) this set is finite and hence bounded for any given $m$. On the other hand, it is well-known that for $m\gg 0$ the line bundle $\det (f_*\omega_{X/C}^m)$ is ample, but we only need that it is not trivial and hence has a non-zero degree.

Now assume that $C$ admits an endomorphism of degree $>1$, say $\sigma:C\to C$. Then the base change $f_\sigma:X_\sigma\to C$ of any admissible family $f:X\to C$ is still admissible, but
$$
\deg ({f_\sigma}_*\omega_{X_\sigma/C}^m) = \deg\sigma \cdot \deg (f_*\omega_{X/C}^m),
$$
which would mean that if non-empty, then $D_m$ could not be bounded, therefore if $C$ admits such an endomorphism, then $D_m$ has to be empty.

If $C$ is an elliptic curve, or a rational curve minus (at most) two points, then it admits such an endomorphism, so they cannot parametrize smooth non-isotrivial families of curves of genus $>1$.

Does it matter whether we consider the stack or the coarse moduli space? I'm interested in knowing the answer in both cases. Could you explain what you mean by "on curvature grounds"?
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FrancescoJul 25 '12 at 23:26

4

@Francesco: Here is the "curvature" argument that @anon probably had in mind (I am assuming that the elliptic curve $E$ is smoothly embedded in the stack Mg). Equip Mg with the Weil-Petersson K\"ahler metric $d_{WP}$; it has negative sectional curvature. Since $E$ is a holomorphic curve, it is also a minimal surface (by Wirtinger inequality), hence, the restriction of $d_{WP}$ to $E$ has smaller curvature that $d_{WP}$ has.Thus,the torus $E$ has a Riemannian metric of negative curvature. This contradicts, say, Gauss-Bonnet formula.
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MishaJul 31 '12 at 14:04